Riemann sum of improper integral over real line

Question

On a physics problem sheet I had to use the following identity:
$$\lim_{a\downarrow0}\sum_{j \in \mathbb{Z}}af(ja) = \int_{-\infty}^{\infty}f(x)dx,$$
where $f: \mathbb{R} \to\mathbb{R}$. Under what conditions on $f$ does this hold? This looks like a Riemann sum where instead of a compact interval the whole real line is partitioned into intervals of length $a$. I've tried to play a little bit around with the definition of the improper integral as the limit of finite integrals but I am stuck. I suppose that $\sum_{j \in \mathbb{Z}}$ means $\lim_{z \to \infty} \sum_{j=-z}^z,$ so the improper integral should be read as taking the Cauchy principal value.

Answer 1

The improper integral cannot always be obtained as a limit of Riemann or Riemann-like sums. This is basically a question of whether or not iterated double limits exist and/or are equal. It depends on the nature of $f$. If $f$ is monotone then it usually works out as discussed here.

As a counterexample, take a function $f$ where $f(k) = 1$ at any integer $k$ and $f(x) = 0$ otherwise. Here we have $$\int_0^\infty f(x) \, dx = 0$$

However, for the sequence $a_n = 2^{-n},$

$$\lim_{a_n \to 0+} \sum_{j=1}^{\infty} a_nf(ja_n) = \lim_{n \to \infty} \lim_{m \to \infty} \frac{1}{2^n} \sum_{j=1}^{m2^n} f(j/2^n) = +\infty$$